Palindrome Partitioning

1. Title

2. Http address

https://leetcode.com/problems/palindrome-partitioning/

3. The question

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s = "aab",
Return

  [
    ["aa","b"],
    ["a","a","b"]
  ]

4. My code (AC)

 1     // Accepted
 2     public List<List<String>> partition(String s) {
 3         List<List<String>> result = new ArrayList<List<String>>();
 4         if( s == null || s.length() <= 0)
 5             return result;
 6         int len = s.length();
 7         String subStr= "";
 8         for(int i = 1; i <= len ;i++)
 9         {
10             List<String> subRes = new ArrayList<String>();
11             subStr = s.substring(0, i);
12             if( !isPalindrome(subStr) )
13             {
14                 continue;
15             }
16             subRes.add(subStr);
17             partitionHlper(s,i,subRes,result);
18         }
19        return result;
20     }
21 
22     private void partitionHlper(String s, int begin, List<String> subRes,
23             List<List<String>> result) {
24         
25         int len = s.length();
26         if( begin >= len)
27         {
28             result.add(new ArrayList<String>(subRes));
29             return;
30         }
31         
32         String subStr ="";
33         for(int i = begin + 1; i <= len; i++)
34         {
35             subStr = s.substring(begin, i);
36             if( !isPalindrome(subStr) )
37             {
38                 continue;
39             }
40             subRes.add(subStr);
41             partitionHlper(s,i,subRes,result);
42             subRes.remove(subRes.size() - 1);
43         }
44         
45     }
46 
47     private boolean isPalindrome(String str) {
48         // TODO Auto-generated method stub
49         if( str == null || str.length() <=1)
50             return true;
51         int len = str.length();
52         int i = 0, j = len -1;
53         while( i < j)
54         {
55             if( str.charAt(i) != str.charAt(j))
56             {
57                 return false;
58             }
59             i++;
60             j--;
61         }
62         return true;
63     }